SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

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1 Chapter Practice Dicsclaimer: The actual eam is different. On the actual eam ou must show the correct reasoning to receive credit for the question. SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Find the absolute etreme values of the function on the interval. 1) g() = , 3 1) ) f(θ) = sin θ + π, 0 θ 7π Write answers in terms of Pi (not a decimal approimation) ) Find the absolute etreme values of the function on the interval. 3) f() = /3, ) Determine all critical points for the function. ) f() = ) ) f() = ) 6) f() = - 1 6) 7) = ) Find the etreme values of the function and where the occur. 8) = ) 9) = ) 10) = ) 11) = ) 1) = ) 13) = ) 1) = ) 1

2 Find the derivative at each critical point and determine the local etreme values. 1) = /3( - ); 0 1) 16) = 9-16) Find the value or values of c that satisf the equation for the function and interval. f(b) - f(a) b - a = f (c) in the conclusion of the Mean Value Theorem 17) f() = + 3 +, [-3, ] 17) 18) f() = +, [6, 9] 18) Determine whether the function satisfies the hpotheses of the Mean Value Theorem for the given interval. 19) f() = 1/3, -, 19) Determine whether the function satisfies the hpotheses of the Mean Value Theorem for the given interval. Wh? 0) f() = 1/3, -1, 0) Determine whether the function satisfies the hpotheses of the Mean Value Theorem for the given interval. 1) g() = 3/, 0, 1) Show that the function has eactl one zero in the given interval. ) f() = , (-, 0). ) 3) r(θ) = cot θ + 1 θ + 3, (0, π). 3) Find all possible functions with the given derivative. ) = 3 - ) ) = 68 ) 6) = 7 t 6) 7) = t - t 7) 8) = csc 3θ 8)

3 Find the function with the given derivative whose graph passes through the point P. 9) f () = + 6, P(0, 19) 9) 30) g () = 1 +, P(-, ) 30) 31) g () = 1 +, P(-, ) 31) Using the derivative of f() given below, determine the intervals on which f() is increasing or decreasing. 3) f () = 1/3( - 8) 3) Find the largest open interval where the function is decreasing. 33) = ) Identif the function's local and absolute etreme values, if an, saing where the occur. Show reasoning, a graph is not proof. 3) f() = ) 3) f(r) = (r - ) 3 3) Use the graph of the function f() to locate the local etrema and identif the intervals where the function is concave up and concave down. 36) 36) - - 3

4 Graph the equation. Include the coordinates of an local and absolute etreme points and inflection points. 37) = ) ) = ) 39) = )

5 0) = 17-0) 1) = 7-1) Sketch the graph and show all local etrema and inflection points. ) = )

6 3) = ) Graph the rational function. ) = ) Solve the problem. ) From a thin piece of cardboard 0 in. b 0 in., square corners are cut out so that the sides can be folded up to make a bo. What dimensions will ield a bo of maimum volume? What is the maimum volume? Round to the nearest tenth, if necessar. ) 6) A compan is constructing an open-top, square-based, rectangular metal tank that will have a volume of 61 ft3. What dimensions ield the minimum surface area? Round to the nearest tenth, if necessar. 6) 6

7 7) A rectangular sheet of perimeter cm and dimensions cm b cm is to be rolled into a clinder as shown in part (a) of the figure. What values of and give the largest volume? 7) 8) Determine the dimensions of the rectangle of largest area that can be inscribed in a semicircle of radius 3. 8) Use Newton's method to estimate the requested solution of the equation. Start with given value of 0 and then give as the estimated solution. 9) = 0; 0 = 1; Find the right-hand solution. 9) 0) = 0; 0 = -1; Find the one real solution. 0) Find an antiderivative of the given function. 1) + 3 1) ) ) 3) ) ) ) ) 3 cos 9 ) 6) cos 7 6) 7) - csc 7) 7

8 Solve the initial value problem. 8) d d = -3/, (1) = 6 8) 9) d d = 1 +, (9) = -3 9) 60) dv dt = 1 csc t cot t, v 3π = - 60) 61) dr dt = 9t + sec t, r(-π) = 61) 6) dr dt = t + sec t, r(-π) = - 6) 63) d d = - 7, (0) = 8, (0) = 63) 6) d 3 d3 = ; (0) = -3, (0) = 7, (0) = 6) 8

9 Answer Ke Testname: 61CHP 1) absolute maimum is 1 at = ; absolute minimum is 0 at 3 and 0 at = ) absolute maimum is 1 at θ = 0; absolute minimum is -1 at θ = π 3) absolute maimum is at = -7 ; absolute minimum is 0 at = 0 ) = - and = ) = 0 and = 6 6) = 1 7) = 0 and = 8) Local maimum at (0, 1), local minimum at (, -3). 9) Local maimum at (-, 18), local minimum at (, -1). 10) Local maimum at (0, -1). 11) The minimum value is - 1 at = -1. The maimum value is 1at = 1. 1) None 13) None 1) The minimum is 8 at = 0. 1) Critical Pt. derivative Etremum Value = 0 = 1 Undefined 0 local ma minimum ) Critical Pt. derivative Etremum Value = 0 = 9 0 undefined min min 0 0 = 36 0 local ma ) ) ) No 0) No 1) Yes ) The function f() is continuous on the open interval (-, 0). Also, f() approaches - as approaches -, and f() approaches as approaches 0 from the left. Since f() is continuous and changes sign along the interval, it must have at least one root on the interval. The first derivative of f() is f () = 3-1, which is everwhere positive on (-, 0). Thus, f() has a single root on 3 (-, 0). 9

10 Answer Ke Testname: 61CHP 3) The function r(θ) is continuous on the open interval (0, π). Also, r(θ) approaches as θ approaches 0 from the right, and r(θ) approaches - as θ approaches π from the left. Since r(θ) is continuous and changes sign along the interval, it must have at least one root on the interval. The first derivative of r(θ) is r (θ) = - csc θ -, which is everwhere negative on (0, π). Thus, r(θ) has a single root θ3 on (0, π). ) - + C ) C 6) 7 t + C 7) t - t + C 8) cot 3θ + C 9) f() = ) g() = ) g() = ) Decreasing on (0, 8); increasing on (-, 0) (8, ) 33) (0, ) 3) local maimum at = -3; local minimum at = 3 3) no local etrema 36) Local minimum at = 1; local maimum at = -1; concave up on (0, ); concave down on (-, 0) 37) absolute minimum: (-1,-8) no inflection points

11 Answer Ke Testname: 61CHP 38) local minimum: (-3, -) local maimum: (3, ) inflection points: (0, 0), (-3 3, -3 3), (3 3, 3 3) ) local minimum: (, -16) local maimum: (1, 11) inflection point:,

12 Answer Ke Testname: 61CHP 0) local minimum: - local maimum: 3, , 17 inflection point: (0, 0) ) local minimum: - local maimum: 1, - 7 1, 7 inflection point: (0, 0)

13 Answer Ke Testname: 61CHP ) Absolute maima: (-1, -6), (1, -6) Local minimum: (0, -7) Inflection points: - 1 3, 3, 1 3, ) Absolute maima: (-, ), (, ) Local minimum: (0, -) Inflection points: - 3, - 9, 3, ) ) 6.7 in. 6.7 in. 6.7 in.; 70.7 in3 13

14 Answer Ke Testname: 61CHP 6) ft ft. ft 7) = 8 cm; = cm 8) h = 3, w = 3 9) 0.3 0) -0. 1) 8 3 3/ + 3 ) 3) ) / ) 1 3 sin 9 6) 7 sin 7 7) cot 8) = 01/ - 1 9) = ) v = - csc t - 61) r = 9 t + tan t π 6) r = t + tan t - - π 63) = ) =